Sodium hydroxide reacts with carbon dioxide as follows: $$ 2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ Which is the limiting reactant when \(1.85 \mathrm{~mol} \mathrm{NaOH}\) and \(1.00 \mathrm{~mol} \mathrm{CO}_{2}\) are allowed to react? How many moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) can be produced? How many moles of the excess reactant remain after the completion of the reaction?

To find the mole ratio, divide the given moles of each reactant by their respective coefficients in the balanced equation: Mole ratio of NaOH = \(\frac{1.85}{2}\) Mole ratio of CO2 = \(\frac{1.00}{1}\)

The reactant with the lowest mole ratio in step 1 will be the limiting reactant. In this case: Mole ratio of NaOH = 0.925 Mole ratio of CO2 = 1.00 Since 0.925 < 1.00, NaOH is the limiting reactant.

Using the stoichiometric ratio from the balanced equation, we can calculate the moles of Na2CO3 produced by the limiting reactant (NaOH): \(moles\, of\, Na_{2}CO_{3} = moles\, of\, NaOH \cdot \frac{1 \,mol\, Na_{2}CO_{3}}{2 \,mol\, NaOH} = 1.85 \,mol\, NaOH \cdot \frac{1}{2} = 0.925 \,mol\, Na_{2}CO_{3}\)

To find the remaining moles of the excess reactant (CO2) after the reaction, we first need to calculate how many moles of CO2 reacted with the limiting reactant (NaOH): \(moles\, of\, CO_{2}\, reacted = moles\, of\, NaOH \cdot \frac{1 \,mol\, CO_{2}}{2 \,mol\, NaOH} = 1.85 \,mol\, NaOH \cdot \frac{1}{2} = 0.925 \,mol\, CO_{2}\) Now, subtract the moles of CO2 reacted from the initial moles of CO2: \(remaining\, moles\, of\, CO_{2} = initial\, moles\, of\, CO_{2} - moles\, of\, CO_{2}\, reacted = 1.00 \,mol\, CO_{2} - 0.925 \,mol\, CO_{2} = 0.075 \,mol\, CO_{2}\) To summarize: 1. The limiting reactant is Sodium hydroxide (NaOH). 2. 0.925 moles of Sodium carbonate (Na2CO3) can be produced. 3. 0.075 moles of Carbon dioxide (CO2) remain after the reaction is complete.